hdu6092 01背包

Rikka with Subset

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 658    Accepted Submission(s): 297


Problem Description
As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Yuta has n positive A1An and their sum is m . Then for each subset S of A , Yuta calculates the sum of S .

Now, Yuta has got 2n numbers between [0,m] . For each i[0,m] , he counts the number of i s he got as Bi .

Yuta shows Rikka the array Bi and he wants Rikka to restore A1An .

It is too difficult for Rikka. Can you help her?  
 

 

Input
The first line contains a number t(1t70) , the number of the testcases.

For each testcase, the first line contains two numbers n,m(1n50,1m104) .

The second line contains m+1 numbers B0Bm(0Bi2n) .
 

 

Output
For each testcase, print a single line with n numbers A1An .

It is guaranteed that there exists at least one solution. And if there are different solutions, print the lexicographic minimum one.
 

 

Sample Input
2
2 3
1 1 1 1
3 3
1 3 3 1
 

 

Sample Output
1 2
1 1 1
 
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e4+88;
typedef long long LL;
LL dp[N],B[N];
int a[N];
int main(){
    int n,m,T;
    for(scanf("%d",&T);T--;){
    memset(dp,0,sizeof(dp));
    memset(a,0,sizeof(a));
    scanf("%d%d",&n,&m);
    for(int i=0;i<=m;++i) scanf("%I64d",&B[i]);
    dp[0]=1;
    for(int i=1;i<=m;++i){
        if(dp[i]==B[i]) continue;
        a[i]=B[i]-dp[i];
        for(int j=1;j<=a[i];++j) for(int k=m;k>=i;--k) dp[k]+=dp[k-i];
    }
    int i;
    for(i=1;i<=m;++i) if(a[i]--) {printf("%d",i);break;}
    for(;i<=m;++i) while(a[i]--) printf(" %d",i);
    puts("");
    }
}

 

posted @ 2017-08-08 21:11  Billyshuai  阅读(116)  评论(0编辑  收藏  举报